The average power generated by the force is approximately \( 48364.7 \, \text{W} \).
The average power (\(P_{\text{avg}}\)) can be calculated using the formula:
\[ P_{\text{avg}} = \frac{W}{\Delta t} \]
where:
- \( W \) is the work done,
- \( \Delta t \) is the time interval.
The work done is given by:
\[ W = F \cdot d \]
where:
- \( F \) is the force,
- \( d \) is the distance.
The force can be found using Newton's second law:
\[ F = m \cdot a \]
Given that the car maintains an acceleration (\( a \)) of \( +4 \, \text{m/s}^2 \) for \( 5.5 \, \text{s} \), and the mass (\( m \)) of the car is \( 1.10 \times 10^3 \, \text{kg} \), we can find the force (\( F \)):
\[ F = m \cdot a \]
\[ F = (1.10 \times 10^3 \, \text{kg}) \cdot (+4 \, \text{m/s}^2) \]
Now, calculate \( F \):
\[ F = 4.40 \times 10^3 \, \text{N} \]
Now, the work done (\( W \)) is:
\[ W = F \cdot d \]
The distance (\( d \)) can be found using the kinematic equation:
\[ d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \]
Since the car starts from rest (\( v_i = 0 \)), the equation simplifies to:
\[ d = \frac{1}{2} a \cdot t^2 \]
Substitute the values:
\[ d = \frac{1}{2} \cdot (+4 \, \text{m/s}^2) \cdot (5.5 \, \text{s})^2 \]
Now, calculate \( d \):
\[ d = 60.375 \, \text{m} \]
Now, substitute \( F \) and \( d \) into the work formula:
\[ W = F \cdot d \]
\[ W = (4.40 \times 10^3 \, \text{N}) \cdot (60.375 \, \text{m}) \]
Now, calculate \( W \):
\[ W = 2.659 \times 10^5 \, \text{J} \]
Now, substitute \( W \) and \( \Delta t \) into the power formula:
\[ P_{\text{avg}} = \frac{W}{\Delta t} \]
\[ P_{\text{avg}} = \frac{2.659 \times 10^5 \, \text{J}}{5.5 \, \text{s}} \]
Now, calculate \( P_{\text{avg}} \):
\[ P_{\text{avg}} \approx 48364.7 \, \text{W} \]
Therefore, the average power generated by the force is approximately \( 48364.7 \, \text{W} \).